Question

Suppose f has a maximum or minimum value at $x=c$. If f is differentiable at $x=c$, what must be true of $f\text{'}\left(c\right)$and why?

Short answer

For $f\text{'}\left(c\right)$to be true, we need to get $f^{\text{'}}\left(c\right)\ge 0,f^{\text{'}}\left(c\right)\le 0$. Thus, $f\text{'}\left(c\right)=0$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

f is differentiable at .

Step 2. Prove if f  is differentiable at x=c,f'c=0.

Suppose a function f has a local maximum at some point $x=c$.

Then, the value $f\left(c\right)$is greater than or equal to all other nearby $f\left(x\right)$values.

If f has minimum at $x=c$. Then, the value $f\left(c\right)$is less than or equal to all other nearby $f\left(x\right)$values.

This implies that $x=c$ is a critical point of f.

Since $x=c$ is the location of a local maximum of f there is some $\delta >0$such that for all localid="1649873351514" $x=(c-\delta ,c+\delta ),f\left(c\right)\ge f^{\text{'}}\left(x\right)$and thus $f\left(x\right)-f\left(c\right)\le 0$.

In this case where $x\in \left(c,c+\delta \right)$it follows that $x>c$which means that $x-c$is positive. Thus,

$f_{+}^{\text{'}}\left(c\right)=\underset{x\to {\infty }^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$

Similarly,

$f_{-}^{\text{'}}\left(c\right)=\underset{x\to {\infty }^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$

This implies that $f^{\text{'}}\left(c\right)\ge 0,f^{\text{'}}\left(c\right)\le 0$. Thus, $f\text{'}\left(c\right)=0$.

Therefore, we can say f is differentiable at$$\begin{gathered} x=c, \\ f\text{'}\left(c\right)=0 \end{gathered}$$.